Part (a): Determine the General Solution
Find the Eigenvalues of A:
To solve the system \mathbf{x}' = A \mathbf{x}, start by finding the eigenvalues \lambda of matrix A:
\det(A - \lambda I) = 0
Compute A - \lambda I:
A - \lambda I = \begin{bmatrix} -2 - \lambda & 1 \\ 0 & -2 - \lambda \end{bmatrix}
Calculate the determinant:
\det(A - \lambda I) = (-2 - \lambda)(-2 - \lambda) - (1)(0) = (-2 - \lambda)^2
Set the determinant to zero:
(-2 - \lambda)^2 = 0 \implies -2 - \lambda = 0 \implies \lambda = -2
Eigenvalue: \lambda = -2 (repeated eigenvalue of multiplicity 2).
Find the Eigenvector of A:
Solve (A - \lambda I)\mathbf{v} = \mathbf{0}:
(A + 2I)\mathbf{v} = \mathbf{0}
Compute A + 2I:
A + 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}
Set up the equations:
\begin{cases}
0 \cdot v_1 + 1 \cdot v_2 = 0 \implies v_2 = 0 \\
0 \cdot v_1 + 0 \cdot v_2 = 0
\end{cases}
Thus, the eigenvector corresponding to \lambda = -2 is:
\mathbf{v} = \begin{bmatrix} v_1 \\ 0 \end{bmatrix} = v_1 \begin{bmatrix} 1 \\ 0 \end{bmatrix}
Apply Theorem 3.5.1:
According to Theorem 3.5.1, if A has a repeated eigenvalue \lambda and only one linearly independent eigenvector \mathbf{v}, the general solution to \mathbf{x}' = A \mathbf{x} is:
\mathbf{x}(t) = c_1 e^{\lambda t} \mathbf{v} + c_2 e^{\lambda t} (t \mathbf{v} + \mathbf{u})
where \mathbf{u} satisfies:
(A - \lambda I) \mathbf{u} = \mathbf{v}
Find the Generalized Eigenvector \mathbf{u}:
Solve (A - \lambda I) \mathbf{u} = \mathbf{v}:
Since A - \lambda I = A + 2I:
(A + 2I) \mathbf{u} = \mathbf{v}
Compute A + 2I as before:
A + 2I = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}
Set up the equation:
\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
From the first row:
0 \cdot u_1 + 1 \cdot u_2 = 1 \implies u_2 = 1
The second row gives:
0 \cdot u_1 + 0 \cdot u_2 = 0 \implies 0 = 0
Choose u_1 = 0 for simplicity. Thus, the generalized eigenvector is:
\mathbf{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
Construct the General Solution:
Using Theorem 3.5.1, substitute \lambda = -2, \mathbf{v} = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, and \mathbf{u} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}:
\mathbf{x}(t) = e^{-2t} \left( c_1 \mathbf{v} + c_2 (t \mathbf{v} + \mathbf{u}) \right)
Simplify:
\mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{-2t} \left( t \begin{bmatrix} 1 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 1 \end{bmatrix} \right)
The general solution is:
\mathbf{x}(t) = c_1 e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} + c_2 e^{-2t} \begin{bmatrix} t \\ 1 \end{bmatrix}
where c_1 and c_2 are constants determined by initial conditions.